Answer
$A^{-1}=\left[\begin{array}{rrr}
{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\\\
{\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\\\
{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$
Work Step by Step
To find the inverse of an $n$ by $n$ nonsingular matrix $A,$ proceed as follows:
STEP 1: Form the matrix $\left[A|I_{n}\right]$
STEP 2: Transform the matrix $\left[A|I_{n}\right]$ into reduced row echelon form.
STEP 3: The reduced row echelon form of $\left[A|I_{n}\right]$ will contain
the identity matrix $I_{n}$ on the left of the vertical bar;
the $n$ by $n$ matrix on the right of the vertical bar is the inverse of $A.$
If the identity matrix can not be obtained on the left, A has no inverse.
----
STEP 1
$\left[A|I_{n}\right] = \left[\begin{array}{rrr|rrr}
{3}&{3}&{1}&{1}&{0}&{0}\\
{1}&{2}&{1}&{0}&{1}&{0}\\
{2}&{-1}&{1}&{0}&{0}&{1}\end{array}\right]\qquad \left(\begin{array}{l}
r_{1}\leftrightarrow r_{2}.\\
.\\
.
\end{array}\right)$
STEP 2
$\rightarrow\left[\begin{array}{rrr|rrr}
{1}&{2}&{1}&{0}&{1}&{0}\\
{3}&{3}&{1}&{1}&{0}&{0}\\
{2}&{-1}&{1}&{0}&{0}&{1}\end{array}\right] \qquad \left(\begin{array}{l}
.\\
R_{2}=r_{2}-3r_{1}.\\
R_{3}=r_{3}-2r_{1}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
{1} & {2} & {1} & {0} & {1} & {0} \\
{0} &{-3}&{-2}&{1} &{-3} &{0}\\
{0} &{-5}&{-1}&{0} &{-2} &{1}\end{array}\right] \qquad \left(\begin{array}{l}
.\\
R_{2}=-2r_{2}+r_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
{1} & {2} & {1} & {0} & {1} & {0} \\
{0} &{1} &{3} &{-2} &{4} &{1}\\
{0} &{-5}&{-1}&{0} &{-2} &{1}\end{array}\right] \qquad\left(\begin{array}{l}
R_{1}=r_{1}-2r_{2}.\\
.\\
R_{3}=r_{3}+5r_{2}.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
{1} & {0} & {-5} & {4} & {-7} & {-2} \\
{0} &{1} &{3} &{-2} &{4} &{1}\\
{0} &{0} &{14} &{-10} &{18} &{6}\end{array}\right] \qquad \left(\begin{array}{l}
.\\
.\\
\div 14.
\end{array}\right)$
$\rightarrow\left[\begin{array}{rrr|rrr}
{1} & {0} & {-5} & {4} & {-7} & {-2} \\
{0} &{1} &{3} &{-2} &{4} &{1}\\
{0} &{0} &{1} &{-5/7} &{9/7} &{3/7}\end{array}\right] \qquad \left(\begin{array}{l}
R_{1}=r_{1}+5r_{3}.\\
R_{2}=r_{2}-3r_{3}.\\
.
\end{array}\right)$
$\rightarrow\left[\begin{array}{ccc|rrr}
{1}&{0}&{0}&{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\\\
{0}&{1}&{0}&{\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\\\
{0}&{0}&{1}&{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$
STEP 3
$A^{-1}=\left[\begin{array}{rrr}
{\displaystyle \frac{3}{7}}&{-\displaystyle \frac{4}{7}}&{\displaystyle \frac{1}{7}}\\\\
{\displaystyle \frac{1}{7}}&{\displaystyle \frac{1}{7}}&{-\displaystyle \frac{2}{7}}\\\\
{-\displaystyle \frac{5}{7}}&{\displaystyle \frac{9}{7}}&{\displaystyle \frac{3}{7}}\end{array}\right]$